推 Xiumpt : 感謝! 09/21 14:36
※ 引述《Xiumpt (進來看熱鬧)》之銘言：
: P(s=1) = 0.5
: y = s + n (n為平均值為0 變異數為1 的Gaussian random variable)
: 求 P(s=1 | y=0.3) = ?
: 我在解的完整要求是 發送端資料S = 1 or -1
: P(s=1) = P(s=-1) = 0.5
: 目前要算的是 P(s=1 | y=0.3) 和 P(s=-1 | y=0.3) 兩個的值
令Phi()為standard normal CDF, 令 d -> 0.
P[s=1 | 0.3<=y<=0.3+d]
= P[s=1, 0.3<=y<=0.3+d] / P[0.3<=y<=0.3+d]
= P[s=1, 0.3<=y<=0.3+d] / (P[s=1, 0.3<=y<=0.3+d] + P[s=-1, 0.3<=y<=0.3+d])
= P[s=1] P[0.3<=y<=0.3+d|s=1] / (P[s=1] P[0.3<=y<=0.3+d|s=1] +
P[s=-1] P[0.3<=y<=0.3+d|s=-1])
= (Phi(-0.7+d) - Phi(-0.7)) / (Phi(-0.7+d) - Phi(-0.7) + Phi(1.3+d) - Phi(1.3))
當d->0時, 分子分母同趨近0. 上下各對d微分, 再令d->0,
原式 = exp(-(-0.7+d)^2/2) / (exp(-(-0.7+d)^2/2) + exp(-(1.3+d)^2/2)
= exp(-0.245) / (exp(-0.245) + exp(-0.845))
= 0.6456563.
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